Answer:
Percent ionic character of HI bond is 4.91%.
Explanation:
<h3>
Given Data:</h3>
Measured Dipole = 0.380D
bond distance = d = 161pm = 1.61*10^-8 cm
<h3>
Calculation:</h3>
% ionic character is determined by following equation:
% ionic= (dipole measured/dipole calculated)*100
Now,
(In above step 3*10^8 is multiplied to convert coulomb into esu)
As,
So,
Now we can % ionic character using above equation:
%ionic=(0.380D/7.728D)*100
% ionic character=4.91%
Answer:
Identifying whether or not an element is an ion is a very simple process. Identify the charge of the element. ... The number of electrons is equal to the atomic number minus the charge of the atom. Refer to an element with either a positive or negative charge as an ion.
Answer:
F
Explanation:
When an atom lose or gain electrons ions are formed.
There are two types of ions. Cation and anion.
Cation:
Cations are formed when atom lose electrons. For example:
X → X⁺ + e⁻
Anion:
It is formed when an atom gain electrons. For example:
X + e⁻ → X⁻
Fluorine is present in group seventeen. It has seven valance electrons. It gain one electron to complete the octet and form anion with charge of -1.
Calcium is alkaline earth metal. It is present in group two it loses two valance electrons and form cation with charge of +2.
Neon is noble gas. It already have complete octet. It can not form ions. To remove the electrons from noble gases very high temperature is required.
Zinc is present in group 12. It usually form cation by losing its two valance electrons but it can also show oxidation state -2.
Answer:
4.034x10^24 atoms
Explanation:
6.7 x 6.023x10^23 = 4.034x10^24 atoms
